Second Order Spectra

BCMB/CHEM 8190

First Order and Second Order NMR Spectra •  The "weak coupling" or "first order" approximation assumes

that, for simple coupled systems, the difference between the Larmor frequencies of the coupled nuclei is large compared to the coupling constant between them:

Δυ >> J•  When the frequency difference approaches the coupling

constant, the spectra are said to be "second order" Δυ ~ J

•  The simple rules presented for first order spectra (multiplicity, number of peaks, peak intensities, chemical shifts of multiplets, values and measurement of coupling constants) do not necessarily apply to second order spectra

•  Importantly, as B0 increases, Larmor frequencies decrease and Δυ decreases, but J is B0 independent, so Δυ J and spectra become second order

•  Spin systems for second order spectra will use letters close in the alphabet (AB, ABC), to indicate similarity in frequencies

Second Order Spectrum Example: Acrylonitrile

For δA=6.3 ppm, δB=6.1 ppm, δC=5.6 ppm:

800 MHz 60 MHz JAB = -2 Hz, ΔυAB = 160 Hz ΔυAB = 12 Hz JAC = 11 Hz, ΔυAC = 560 Hz ΔυAC = 42 Hz JBC = 16 Hz, ΔυBC = 400 Hz ΔυBC = 30 Hz

N≡C

1HC 1HA

1HB

A B C

AMX first order spectrum

•  Shown is an "ideal" first order spectrum (AMX) for acrylonitrile

•  At lower B0 (as Δυ J ) the spectrum becomes second order

•  Are 12 peaks, equal intensities (predicted for first order)

•  In first order spectra, can't discern the sign of the coupling (by looking at the spectra). Not always the case for second order

Field dependence of second order spectra

60 MHz 90 MHz

250 MHz 490 MHz

•  As Δυ J, a number of noticeable effects - peak intensities deviate from what is expected for first order - number of peaks changes (can get peaks for "forbidden" transitions) - signals become unrecognizable with respect to first order expectations

Field dependence of second order spectra simulated spectra

Peak Intensities and Transition Probabilities •  Recall for pairs of spin ½ nuclei, the product wavefunctions (αα, βα, αβ, and ββ), in the limit of first order spectra (AX), are good solutions to Schrödinger's equation

- they form a complete orthonormal set •  Can write wavefunctions for more complex Hamiltonians by

making linear combinations of all members of the set (this is how second order spectra are treated)

ψ = c1αα + c2αβ + c3βα + c4ββ = cjϕ jj∑

•  Recall, the probability of being in some state, ψ, is

•  Thus, the probability (ρ) of (mostly) being in one (j) of our well-defined basis states is

ψ∗ψ

ρ jj ∝ cj∗  cj

•  The transition probability, the probability of starting in one state (k) and ending in another (l), is

ρk→l ∝ ck∗  ck( )  cl∗  cl( )

Peak Intensities and Transition Probabilities •  The transition probability, the probability of starting in one state

(k) and ending in another (l), is ρk→l ∝ ck

∗  ck( )  cl∗  cl( )•  If we know we start in state k (c*kck = 1), then we only need to

solve for the probability that we are going to end up in state l (peak intensities)

•  This requires consideration of the time dependence (for transitions to occur), hence Schrödinger's time dependent equation (time derivative of the wavefunction)

Hψ(t) = −i! d(ψ(t)) dt ψ = cjϕ jj∑

•  So, to see how cl changes with time (in unit time, what is the probability of going from state k to l)

d(cl ) dt = φl H' φk     integrated from t = 0 to 1

•  Last thing: have to square the result to get intensities ρk→l ∝ φl H' φk

2 

Understanding second order spectra •  Pure first order (AX) spectrum: HF - even at low field strength, Larmor frequencies of 1H and 19F are MHz apart (at 2.35 T, 100 MHz 1H and 94 MHz 19F, Δυ ~ 6,000,000 Hz), and the one-bond HF coupling constant is ~500 Hz, so, Δυ >>> J

1H spectrum 19F spectrum

ββ αβ βα αα

E

JAX=0 JAX>0

spin 1 spin 2

spin

1

spin

2

spin

1

spin

2

•  Transition energies when J > 0 (i.e. positive coupling)

- when J > 0, ΔE for transitions to/from the highest energy state (ββ) increases compared to J = 0, ΔE for transitions to from the lowest energy state (αα) decrease compared to J = 0

•  Transition energies when J = 0 (i.e. no coupling)

- for spin 1, αα βα and αβ ββ energies are equal - for spin 2, αα αβ and βα ββ energies are equal

Understanding second order spectra •  Peak intensities for "pure" first order (AX)

spectrum: HF - the intensities of the signals in each multiplet should be identical - we can calculate the probabilities for single quantum (Δm = 1) interconversions, which should, therefore, all be the same

1H spectrum 19F spectrum

ρk→l∝ φl H' φk2 

ρββ→αβ ∝ ββ γB1(Ix1 + Ix2 ) αβ2

=  γB1 ββ (Ix1) αβ + ββ (Ix2 ) αβ⎡⎣ ⎤⎦ 2

=  γB1 1 2 ββ ββ +1 2 ββ αα⎡⎣ ⎤⎦ 2

=  γB1 1 2+ 0[ ] 2

•  For the first order system, we'll write the Hamiltonian as H'=γB1( Ix1 + Ix2 )

- we can leave out the time dependence (rotating frame) - use B1 (rf field) which interacts with the magnetic moment (γIx, assuming B1 acts along x-axis) •  Example: evaluate for ββ αβ transition ( )

(recall Ix α =1 2β     Ix β =1 2α)

•  Evaluate all four transitions: ραα→αβ = ραα→βα = ραβ→ββ = ρβα→ββ

•  What about αα ββ ???

ββ

(αβ+βα)/√2

αα

(αβ-βα)/√2

E

Understanding second order spectra •  Consider a "pure" second order (A2) spectrum (H2, H2O, etc.) - expect single peak (equivalent nuclei have identical chemical shifts) - coupling does not result in peak splitting…....why?. •  Problem: αα, αβ, βα, ββ are not individually good solutions to

Schrödinger's equation for equivalent nuclei - recall, when not first order, must include IxIx and IyIy in the scalar coupling term of the Hamiltonian (first order approximation no longer valid) - also, αβ and βα imply they are distinguishable (can't be for identical nuclei) - αα and ββ are OK •  Solution: use linear combinations of αβ and βα

(αβ +βα) 2       (αβ −βα) 2

ρββ→(αβ+βα ) ∝ ββ γB1(Ix1 + Ix2 ) αβ +βα( ) / 22

ρββ→(αβ−βα ) ∝ ββ γB1(Ix1 + Ix2 ) αβ −βα( ) / 22

•  So, we can then solve for the 4 possible transition probabilities

ρββ→(αα+βα ) ∝ αα γB1(Ix1 + Ix2 ) αβ +βα( ) / 22

ρββ→(αα−βα ) ∝ αα γB1(Ix1 + Ix2 ) αβ −βα( ) / 22

ρk→l ∝ φl H' φk2 

ρββ→(αβ+βα ) ∝ ββ γB1(Ix1 + Ix2 ) αβ +βα( ) / 22

= γB1 ββ Ix1 αβ +βα( ) / 2 +γB1 ββ Ix2 αβ +βα( ) / 22

= γB1 ββ Ix1 αβ( ) / 2 + ββ Ix1 βα( ) / 2 + ββ Ix2 αβ( ) / 2 + ββ Ix2 βα( ) / 2⎡⎣

⎤⎦

2

= γB1 (1 2) 2 ββ ββ + (1 2) 2 ββ αα + (1 2) 2 ββ αα + (1 2) 2 ββ ββ⎡⎣

⎤⎦2

= γB1 (1 2) 2 + 0+ 0+ (1 2) 2⎡⎣

⎤⎦2

= γB1 1 2⎡⎣

⎤⎦2

=1 2 (γB1)2

Understanding second order spectra •  Solve for one of the transitions

ρββ→(αβ+βα ) ∝1 2 (γB1)2

ρββ→(αβ−βα ) = 0

ραα→(αβ+βα ) ∝1 2 (γB1)2

ραα→(αβ−βα ) = 0

•  All solutions

ββ

(αβ+βα)/√2

αα

(αβ-βα)/√2

E

•  Result - two transitions of equal energy and equal probability (i.e. one peak) - two transitions of with different energies, but zero probability (not allowed)

"Intermediate" behavior of 2nd order spectra

H2O AB HF

outer lines are not observed, but couplings still exist

•  AB systems are "intermediate" between AX and A2 - as Δυ J an AX system becomes AB - inner peaks of doublets more intense than outer peaks ("roof effect") - J can still be measured as distance between peak centers - chemical shifts are not center of doublets, but weighted averages of peak intensities ("center of gravity"), and can be determined knowing

€

δA −δB = (δ1 −δ4 )(δ2 −δ3)

ABX systems

X

J45? O

HO OH

CBaseH5'

H5

OHH4

φ

- software: S.A. Smith, J. Magn. Res., 166, 75 (1994) M.Veshtort, R.G.Griffin, J Magn. Res., 178, 248-282 (2006)

- for 'X', can't necessarily deduce coupling constant from peak separation - example: coupling of H5 (A), H5' (B) and H4 (X) in ribose rings

- may consider using Karplus relationship to deduce torsion angle from J45 and J45' - splitting in ABX spectrum in X signal is equal to (J45+J45')/2 - can't conclude J45 = J45'

•  ABX systems raise additional complications

•  Helpful to simulate spectra